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At 310   K , the solubility of CaF 2 in water is 2 . 34 × 10 - 3   g / 100   mL . The solubility product of CaF 2 is ---- × 10 - 8 mol / L 3 (nearest integer). (Given molar mass : CaF 2 = 78   g   mol - 1 )

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The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol K sp . Solubility of CaF 2 = S   mole / / L S = 2 . 34 × 10 - 3 0 . 1 × 78 = 2 . 34 78 × 10 - 2 = 3 × 10 - 4   mol / L K sp CaF 2 = 4   S 3 = 4 3 × 10 - 4 3 = 108 × 10 - 12 = 0 . 0108 × 10 - 8 mol / L 3

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